题意：
多n个圆的面积并。
思路：
发现和求2个圆的完全不一样，具体请参考
SPOJ 8073 The area of the union of circles（计算几何の圆并）（CIRU）
（用格林公式搞真是跪烂了。。。。
没有仔细看细节，当成板子好了(我最菜.jpg
将代码写成了自己熟悉的风格。
以及双倍经验题:SPOJ VCIRCLES
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/* *********************************************** Author :111qqz Created Time :2017年10月11日 星期三 19时53分30秒 File Name :ciru.cpp ************************************************ */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <cmath> #include <cstdlib> #include <ctime> #define PB push_back #define fst first #define sec second #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define ms(a,x) memset(a,x,sizeof(a)) typedef long long LL; #define pi pair < int ,int > #define MP make_pair using namespace std; const double eps = 1E-8; const int dx4[4]={1,0,0,-1}; const int dy4[4]={0,-1,1,0}; const int inf = 0x3f3f3f3f; const double PI = acos(-1.0); const int N =1e3+7; inline int dblcmp( double d) { return d<-eps?-1:d>eps;} struct point { double x,y; point(){} point(double x,double y):x(x),y(y){} void input(){scanf("%lf%lf",&x,&y);} double angle(){ return atan2(y,x);} point operator + (const point &rhs)const{ return point(x+rhs.x,y+rhs.y);} point operator - (const point &rhs)const{ return point(x-rhs.x,y-rhs.y);} point operator * (double t)const{ return point(t*x,t*y);} point operator / (double t)const{ return point(x/t,y/t);} double length() const { return sqrt(x*x+y*y);}; point unit()const { double l = length();return point(x/l,y/l); } }; double cross (const point a,point b){ return a.x*b.y-a.y*b.x ;} double dist(const point p1,point p2) { return (p1-p2).length();} point rotate(point p,double angle,point o = point(0,0)) { point t = p-o; double x = t.x * cos(angle) - t.y*sin(angle); double y = t.y * cos(angle) + t.x*sin(angle); return point (x,y)+o; } struct region{ double st,ed; region(){} region(double st,double ed):st(st),ed(ed){} bool operator < (const region & rhs)const { if (dblcmp(st-rhs.st)) return st<rhs.st; return ed<rhs.ed; } }; struct circle{ point c; double r; vector <region>reg; circle(){} circle(point c,double r):c(c),r(r){} void input() { c.input(); scanf("%lf",&r); } void add(const region &r){ reg.PB(r);} bool contain (const circle & cir)const{ return dblcmp(dist(cir.c,c)+cir.r-r)<=0;} bool interect (const circle & cir)const{ return dblcmp(dist(cir.c,c)-cir.r-r)<0;} }; double sqr( double x){ return x*x;} void intersection(circle cir1,circle cir2,point &p1,point &p2) { double l = dist(cir1.c,cir2.c); double d = (sqr(l)-sqr(cir2.r) + sqr(cir1.r))/(2*l); double d2 = sqrt(sqr(cir1.r)-sqr(d)); point mid = cir1.c + (cir2.c-cir1.c).unit() * d; point v = rotate(cir2.c-cir1.c,PI/2).unit()*d2; p1 = mid + v,p2 = mid -v; } point calc(const circle &cir,double angle) { point p = point (cir.c.x+cir.r,cir.c.y); return rotate(p,angle,cir.c); } circle cir[N]; bool del[N]; int n; double solve() { double ans = 0 ; for ( int i = 0 ; i < n ; i++){ for ( int j = 0 ; j < n ; j++) if (!del[j]){ if (i==j) continue; if (cir[j].contain(cir[i])) { del[i] = true; break; } } } for ( int i = 0 ; i < n ; i++) if (!del[i]){ circle &mc = cir[i]; point p1,p2; bool flag = false; for ( int j = 0 ; j < n ; j++) if (!del[j]){ if (i==j) continue; if (!mc.interect(cir[j])) continue; flag = true; intersection(mc,cir[j],p1,p2); double rs = (p2-mc.c).angle(),rt = (p1-mc.c).angle(); if (dblcmp(rs)<0) rs+=2*PI; if (dblcmp(rt)<0) rt+=2*PI; if (dblcmp(rs-rt)>0) mc.add(region(rs,PI*2)),mc.add(region(0,rt)); else mc.add(region(rs,rt)); } if (!flag) { ans += PI*sqr(mc.r); continue;} sort(mc.reg.begin(),mc.reg.end()); int cnt = 1; for ( int j = 1 ; j < mc.reg.size() ; j++) { if (dblcmp(mc.reg[cnt-1].ed - mc.reg[j].st)>=0) mc.reg[cnt-1].ed = max(mc.reg[cnt-1].ed,mc.reg[j].ed); else mc.reg[cnt++] = mc.reg[j]; } mc.add(region()); mc.reg[cnt]=mc.reg[0]; for ( int j = 0 ; j < cnt ; j++) { p1 = calc(mc,mc.reg[j].ed); p2 = calc(mc,mc.reg[j+1].st); ans +=cross(p1,p2)/2; double angle = mc.reg[j+1].st - mc.reg[j].ed; if (dblcmp(angle)<0) angle+=2*PI; ans+=0.5*sqr(mc.r)*(angle-sin(angle)); } } return ans; } int main() { #ifndef ONLINE_JUDGE freopen("./in.txt","r",stdin); #endif scanf("%d",&n); for ( int i = 0 ; i < n ; i++) cir[i].input(); printf("%.3f\n",solve()); #ifndef ONLINE_JUDGE fclose(stdin); #endif return 0; } |
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/* *********************************************** Author :111qqz Created Time :2017年10月11日 星期三 19时53分30秒 File Name :ciru.cpp ************************************************ */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <cmath> #include <cstdlib> #include <ctime> #define PB push_back #define fst first #define sec second #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define ms(a,x) memset(a,x,sizeof(a)) typedef long long LL; #define pi pair < int ,int > #define MP make_pair using namespace std; const double eps = 1E-8; const int dx4[4]={1,0,0,-1}; const int dy4[4]={0,-1,1,0}; const int inf = 0x3f3f3f3f; const double PI = acos(-1.0); const int N =1e3+7; inline int dblcmp( double d) { return d<-eps?-1:d>eps;} struct point { double x,y; point(){} point(double x,double y):x(x),y(y){} void input(){scanf("%lf%lf",&x,&y);} double angle(){ return atan2(y,x);} point operator + (const point &rhs)const{ return point(x+rhs.x,y+rhs.y);} point operator - (const point &rhs)const{ return point(x-rhs.x,y-rhs.y);} point operator * (double t)const{ return point(t*x,t*y);} point operator / (double t)const{ return point(x/t,y/t);} double length() const { return sqrt(x*x+y*y);}; point unit()const { double l = length();return point(x/l,y/l); } }; double cross (const point a,point b){ return a.x*b.y-a.y*b.x ;} double dist(const point p1,point p2) { return (p1-p2).length();} point rotate(point p,double angle,point o = point(0,0)) { point t = p-o; double x = t.x * cos(angle) - t.y*sin(angle); double y = t.y * cos(angle) + t.x*sin(angle); return point (x,y)+o; } struct region{ double st,ed; region(){} region(double st,double ed):st(st),ed(ed){} bool operator < (const region & rhs)const { if (dblcmp(st-rhs.st)) return st<rhs.st; return ed<rhs.ed; } }; struct circle{ point c; double r; vector <region>reg; circle(){} circle(point c,double r):c(c),r(r){} void input() { c.input(); scanf("%lf",&r); } void add(const region &r){ reg.PB(r);} bool contain (const circle & cir)const{ return dblcmp(dist(cir.c,c)+cir.r-r)<=0;} bool interect (const circle & cir)const{ return dblcmp(dist(cir.c,c)-cir.r-r)<0;} }; double sqr( double x){ return x*x;} void intersection(circle cir1,circle cir2,point &p1,point &p2) { double l = dist(cir1.c,cir2.c); double d = (sqr(l)-sqr(cir2.r) + sqr(cir1.r))/(2*l); double d2 = sqrt(sqr(cir1.r)-sqr(d)); point mid = cir1.c + (cir2.c-cir1.c).unit() * d; point v = rotate(cir2.c-cir1.c,PI/2).unit()*d2; p1 = mid + v,p2 = mid -v; } point calc(const circle &cir,double angle) { point p = point (cir.c.x+cir.r,cir.c.y); return rotate(p,angle,cir.c); } circle cir[N]; bool del[N]; int n; double solve() { double ans = 0 ; for ( int i = 0 ; i < n ; i++){ for ( int j = 0 ; j < n ; j++) if (!del[j]){ if (i==j) continue; if (cir[j].contain(cir[i])) { del[i] = true; break; } } } for ( int i = 0 ; i < n ; i++) if (!del[i]){ circle &mc = cir[i]; point p1,p2; bool flag = false; for ( int j = 0 ; j < n ; j++) if (!del[j]){ if (i==j) continue; if (!mc.interect(cir[j])) continue; flag = true; intersection(mc,cir[j],p1,p2); double rs = (p2-mc.c).angle(),rt = (p1-mc.c).angle(); if (dblcmp(rs)<0) rs+=2*PI; if (dblcmp(rt)<0) rt+=2*PI; if (dblcmp(rs-rt)>0) mc.add(region(rs,PI*2)),mc.add(region(0,rt)); else mc.add(region(rs,rt)); } if (!flag) { ans += PI*sqr(mc.r); continue;} sort(mc.reg.begin(),mc.reg.end()); int cnt = 1; for ( int j = 1 ; j < mc.reg.size() ; j++) { if (dblcmp(mc.reg[cnt-1].ed - mc.reg[j].st)>=0) mc.reg[cnt-1].ed = max(mc.reg[cnt-1].ed,mc.reg[j].ed); else mc.reg[cnt++] = mc.reg[j]; } mc.add(region()); mc.reg[cnt]=mc.reg[0]; for ( int j = 0 ; j < cnt ; j++) { p1 = calc(mc,mc.reg[j].ed); p2 = calc(mc,mc.reg[j+1].st); ans +=cross(p1,p2)/2; double angle = mc.reg[j+1].st - mc.reg[j].ed; if (dblcmp(angle)<0) angle+=2*PI; ans+=0.5*sqr(mc.r)*(angle-sin(angle)); } } return ans; } int main() { #ifndef ONLINE_JUDGE freopen("./in.txt","r",stdin); #endif scanf("%d",&n); for ( int i = 0 ; i < n ; i++) cir[i].input(); printf("%.5f\n",solve()); #ifndef ONLINE_JUDGE fclose(stdin); #endif return 0; } |
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